I gave the example not to teach you, but so that we could get away from terminology and get to meaning. Look at the question I posed: "how many arrows would it take pointing from the student to the teacher to illustrate every possible assignment?" So, if you have two students Alice and Eve, and two teachers Bob and Carl, there are four arrows: Alice->Bob, Alice->Carl, Eve->Bob, and Eve->Carl. How many arrows are there if there are no teachers and no students?
The number of arrows is |students x teachers|, every student paired with every teacher. From this you get a^0 = 0, but it also follows that a^b = b^a. This defines multiplication, not exponentiation. The problem is that you cannot illustrate the assigment where Eve is not paired with Bob using an arrow, but only by not using a one.
I gave the example not to teach you, but so that we could get away from terminology and get to meaning. Look at the question I posed: "how many arrows would it take pointing from the student to the teacher to illustrate every possible assignment?" So, if you have two students Alice and Eve, and two teachers Bob and Carl, there are four arrows: Alice->Bob, Alice->Carl, Eve->Bob, and Eve->Carl. How many arrows are there if there are no teachers and no students?